Here is the way I am using to return duplicate elements.. But I am facing most dangerous performance issues like browser close etc when my array have large number of items with long texts..
var arr = [9, 9, 111, 2, 3, 4, 4, 5, 7];
var sorted_arr = arr.sort();
var results = [];
for (var i = 0; i < arr.length - 1; i++) {
if (sorted_arr[i + 1] == sorted_arr[i]) {
results.push(sorted_arr[i]);
}
}
alert(results);
Please suggest me a best way of doing this
i don't get exactly what you want, but if you need to return duplicates you could use a cache object. this works with number or string or whatever.
var arr = [9, 9, 111, 2, 3, 4, 4, 5, 7];
var cache = {};
var results = [];
for (var i = 0, len = arr.length; i < len; i++) {
if(cache[arr[i]] === true){
results.push(arr[i]);
}else{
cache[arr[i]] = true;
}
}
console.log(results);//returns an array with 9 and 4
Of course you can do other things like deleting multiple items etc. etc.
EDIT - i've written a blog entry on how to remove duplicates from an array
We have array filter, and also indexOf and lastIndexOf,
so you can return the duplicates without doing the sort.
var results, arr = [9, 9, 111, 2, 3, 4, 4, 5, 4, 7];
results = arr.filter(function(itm, i){
return arr.lastIndexOf(itm)== i && arr.indexOf(itm)!= i;
});
console.log(results);
/* returned value: (Array)
9,4
*/
Assuming Nicola's solution doesn't work for you (since it uses about as much memory as the original solution: two elements stored per element in the input, worst-case), you can use the slower process of repeatedly searching your input.
This requires the Array.indexOf method from ECMAScript 5. A lot of browsers have it. For alternatives, see How do I check if an array includes an object in JavaScript?.
var arr = [9, 9, 111, 2, 3, 4, 4, 5, 7];
var results = [];
for (var i = 0, len = arr.length - 1; i < len; i++) {
if((results.indexOf(arr[i]) == -1) && (arr.indexOf(arr[i], i + 1) != -1)) {
results.push(arr[i]);
}
}
console.log(results);
This uses no more memory than the input arr plus the output results, but it's an O(N^2) algorithm and doesn't have to modify arr.
Your method relies on a sort, which may or may not be one reason you run out of space/time.
The canonical way to remove duplicates is to keep a hash map of the keys (an object in JS). The object keys you get back won't necessarily be in the order you want; you don't specify if you want the results ordered as well, but they are now.
You could null out the original array, since you no longer require it; when it gets collected is up to the JS engine though.
You could remove duplicates "in place" by keeping a "current index" into the sorted array, and increment it only when you move a non-duplicated element "down" from the counter index, then truncate the array that you return.
Combining the last two techniques should mean that in general you'll only have a single array with a valid reference.
Edit Example. Setting length explicitly, as .slice() creates a new array.
var have = {};
var arr = [9, 9, 111, 2, 3, 4, 4, 5, 7];
arr = arr.sort();
for (var rIdx = 0, i = 0; i < arr.length; i++) {
if (have[arr[i]]) {
arr[rIdx++] = arr[i];
} else {
have[arr[i]] = true;
}
}
arr.length = rIdx;
console.log(arr);
Filter the array using array.filter(), including elements with a different first occurrence index and acting as the last occurrence in the array.
const returnDuplicates = arr => arr.filter((item, index) => arr.indexOf(item) !== index && arr.lastIndexOf(item) === index);
const array = [1, 1, 1, 1, 2, 2, 2, true, true, "hello", "hello"];
console.log(returnDuplicates(array));